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Unformatted text preview: 3.210 A rigid tank is disdded into two rooms by a membrane, both containing water,
shown in Fig. P3.210. Room A is at 200 lsPa= v = 0.5 n131'kg, VA =1 m3, and room B contains 3.5 kg at 0.5 MP'a= 400°C. The membrane nowr ruptures and heat
transfer takes place so the water comes to a uniform state at 100°C. Find the heat
transfer during the process. Solution: C.V.: Both rooms A and B in tank. A
Continuityr Eq.: n12=mAl +1nBl ; Energy Eq': ”12% ' ml13.1mm ' mBluBl : 1Q: ' 1W2 State 1A: (P= 1:] Table 13.1.2= mAl = VAJ'VAI =11"0.5 = 2 kg V—Vf 0.5  0.001061 3.41— % ‘ 0.88467 ‘0'564 “A1 = uf+ x ufg = 504.4? + 0.564 x 2025.02 =1646.6 kakg State 1B: Table B.1.3= VB . . _ _ 3 _
Process constant total solume. Vtot — VA + VB — 3.16 m and 1W2 — 0 _ _ _ _ _ 3
1— 0.6173, “Bl — 2963.2= VB — 111311131 — 2.16 m m2=rnAl+1nBl=i5kg =l> VEZVto n12=0.57461n31’kg State 2: T2 , ‘02 => Table 13.1.1 twophase as v2 ‘1 vg _ V2 — Vf _ 0.5746 — 0.001044 _
32 — vfg _ 1.67185 _ u2 =uf+xufg=418.91+0.343 >< 208158: 1134.95 kakg 0.343 , Heat transfer is ﬂow the energy equation
1'32 2 ”12% ' ”1141111141 ' 111311131 = (5.5 x1134.95  2 X16466 — 3.5 x 2953.2) kg x mkg
=7421 kJ 3.212 The cylinder volume below the constant loaded piston has two compartments A
and B ﬁlled with water. A has 0.5 kg at 200 kPa, 150°C and B has 400 kPa with a
qualityr of 50% and a volume of 0.1 m3. The valve is opened and heat is
transferred so the water comes to a uniform state with a total volume of 1.006 1113. a) Find the total mass of water and the total initial volume. b) Find the work in the process c) Find the process heat transfer. Solution:
Take the water in A and B as CV.
Continuity: m2  mlA  “113 = 0 Energy: 1112112  mlallm  111131113 = 102  1W2
Process: P = constant = P1 A if piston ﬂoats (VA positive) i.e. if V2 2> VB = 0.1 m3
State A1: Sup. vap. Table 3.1.3 v = 0.95964 m3."Iig= u = 25 26.9 kJ."kg => V = mv = 0.5 kg X 0.95964 m3.'l(g = 0.42982
State B]: Table 13.1.2 v = (1K) X 0.001084 + x X 0.4625 = 0.2318 m3."kg
=3: m = V."'v = 0.4314 kg 11 = 604.29 + 0.5 X 1949.3 = 1578.9 Iii"kg State 2: 200 lcF'a= v2 = V2_"'n‘1 = 1.006."'0.9314 = 1.0301 m3_"]ig Table 3.1.3 => close to T2 = 200°C and 112 = 2654.4 kJ."kg
So now
V1: 0.42932 + 0.1 = 0.5798 m3, m1 = 0.5 + 0.4314 : 0.9314 kg Since volume at state 2 is larger than initial volume piston goes up and the
pressure then is constant (200 kPa which ﬂoats piston). 1w2 =1 P W = PM, (v2  v1) = 200 kPa (1.006  0.57932) m3 = 35.24 kJ 1Q2 = ”12112 ' mlAulA ' mlBulB +1w2
= 0.9314 >< 2654.4  0.5 X 2526.9  0.4314 >< 1523.9 + 35.24 = 538 kJ 3.215 A tank has a volume of 1 m3 with oxygen at 150C= 300 kPa. Another tank contains 4 kg oxygen at 60W:= 500 kPa. The two tanks are connected by a pipe
and valve which is opened allowing the whole system to come to a single equilibrium state with the ambient at 20°C. Find the ﬁnal pressure and the heat
transfer. CV. Both tanks of constant volume. Continuity Eq.: n12 — mlA — mm = 0 Energy 1361.: 1112112 — 1111251111191 — mlBulB = .1121?2 — 1W2 Process Eq.: V2 = VA + VB = constant= 1W2 = {J
5 IA _P1AVA_ 300 kPa >< 11113 4007k
we ' “1hr RT1A _ 0.2593 kl‘kg—K x 233.15 K _ ' g
mlBRTlB 4 kg 3 0.2593 kl"kg—K 3 333.15 K 3
State 1B: VB — P13 — 500 kPa — 0.6924 m
State 2: (r2. '2 = v2..'m2) v2 = VA + VB = 1 + 0.6924 = 1.6924 613
mzRTg 3.007 kg 3 0.2593 ktItgK 3 293.15 K
P2 = = = 360.3 kPa V2 1.6924 613
Heat transfer ﬁ‘om energy equation
IQQ = 1112112 — 13019111121 — mlBulB = m1A(u2 — 111A) + 11113012 — “13)
= 111va (T2 — TIA) + 1111113 Cv (T2 — T13)
= 4.002 kgx0.662 kJ..1tg—K><(20 —15) K + 4 kgx0.662 kJ."kgK 3(20 450) K
= — 92.65 H 3.221 Two springs with same spring constant are installed in a massless pistonfcylinder
with the outside air at 100 kPa. Ifthe piston is at the bottom, both springs are relaxed and the second spring comes in contact with the piston at 17: 2 n13. The
cylinder (Fig. P3221) contains ammonia initially at —2°C, 1 = 0.13, V: 1 m3,
which is then heated until the pressure ﬁnally reaches 1200 kPa. At what pressure will the piston touch the second spring? Find the ﬁnal temperature, the total work
done by the ammonia and the heat transfer. Solution : State 1: P = 399.7 kPa Table B.2.1 v = 0.00156 + (1133403106 = 0.0419 mgfkg
u = 170.52 + 01324114578 = 319.47 kakg
m = Viv = 1f0.0419 = 23.866 kg At bottom state 0: 0 m3, 100I kPa State 2: v: 2 m3 and online 0.1.2
Final state 3: 120CI kPa, on line segment 2. Slope of line 012: 313331! = (P1  13,9131; =(399.7100)11 = 299.7 kPaf 313 P2 = P1 + (V2  VﬂAPfﬁV = 399.7 + (21)):299'? = 699.4 kPa
State 3: Last line segment has twice the slope. P3 = P; + (V3  V2)2APM.V => V3 = 1172+ (Pg  PﬁfﬂﬁPJ’ﬁV) V3 = 2 + (1200699.4)f599.4 = 2.835 m3 V3 = V1V3Nl = 0.0419x2.8351'1= 0.1188 2} T3 = 51°C 01133—011346
113 = 1334.39 kmcg = 1333 + (140431333) 0123.”, _ 011346 from 3.2.2 1 l
1W3=1W2+2W3=§lpl+PﬂW2V1)+§(P3+P2)W3Vzl = 549.6 + 793.0 = 1342.6 Id The energy equation gives the heat transfer as
1Q3 = ma;3 — 111) + 1W3 = 23.866 kg (1334.39 — 319.47) 1;ng + 1342.5 1:] = 26 758 kJ ...
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 Spring '08
 Sherif

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